Calculating quality factor in a series LCR circuit?

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CrimsonFlash
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Homework Statement


My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
max energy = 0.5 L I2
energy lost per cycle = 0.5 R I2 2π/ω .
..and at resonance he calculates it to be equal to be [tex]\frac{1}{R} \sqrt{L/C} [\tex]<br /> <br /> <h2>Homework Equations</h2><br /> -<br /> <br /> <h2>The Attempt at a Solution</h2><br /> However, this is how I tried to do it:<br /> max energy = 0.5 C V<sup>2</sup><br /> energy lost per cycle = 0.5 V<sup>2</sup> /R x 2π/ω<br /> ...and at resonance this gives a quality factor of [tex]R \sqrt{C/L} [\tex]<br /> <br /> Erm... what da heck?[/tex][/tex]
 
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Latex...y u no work?
 
I'll post the answers here :
Q from the 1st bit: 1/R x (L/C)^(1/2)
Q from my try: R x (C/L)^(1/2)
 
CrimsonFlash said:

Homework Statement


My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
max energy = 0.5 L I2
energy lost per cycle = 0.5 R I2 2π/ω .
..and at resonance he calculates it to be equal to be [tex]\frac{1}{R} \sqrt{L/C} [\tex]<br /> <br /> <h2>Homework Equations</h2><br /> -<br /> <br /> <h2>The Attempt at a Solution</h2><br /> However, this is how I tried to do it:<br /> max energy = 0.5 C V<sup>2</sup><br /> energy lost per cycle = 0.5 V<sup>2</sup> /R x 2π/ω<br /> ...and at resonance this gives a quality factor of [tex]R \sqrt{C/L} [\tex]<br /> <br /> Erm... what da heck?[/tex][/tex]
[tex][tex] <br /> .<br /> <br /> Q = wL/R = 1/wRC, w = 1/sqrt(LC)<br /> Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)<br /> <br /> Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor. <br /> <br /> max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).<br /> What is dissipation in R in terms of V0?<br /> <br /> Take it from there.[/tex][/tex]
 
rude man said:
.

Q = wL/R = 1/wRC, w = 1/sqrt(LC)
Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)

Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor.

max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).
What is dissipation in R in terms of V0?

Take it from there.

Max Energy: 0.5 C (V/ωRC)2
Energy lost per cycle: V2/2R x 2π/ω
...and this gives the correct value for Q.

Thank you.